mixing formulas questions 1-3
Introduction
Presenter: Jim Matt
Objectives:
Review applications of the mixing formula.
Solve examples, with a focus on work color in example seven.
Gain proficiency in using the mixing formula for various brewery problems.
Applications of the Mixing Formula
Versatile tool in the brewery for solving problems such as:
Alcohol correction: Diluting high alcohol beer with water.
Caustic makeup: Creating lower concentration caustic solutions.
Peracetic acid makeup: Preparing sanitizer solutions.
Dilution water calculations.
Heat balance issues.
Multivariable three-component mixing.
Overview of the Mixing Formula
Formula Structure:
A × a + B × b = C × c
Capital letters (A, B, C): quantities to be mixed.
Lowercase letters (a, b, c): corresponding properties (e.g., alcohol by volume).
Example units: liters, milliliters, hectoliters.
Example 1: Alcohol Correction Problem
Problem: Determine final alcohol percentage when mixing high gravity beer and water.
Given:
A = 200 hectoliters (high gravity beer)
a = 7.15% (alcohol by volume)
B = 100 hectoliters (dilution water)
b = 0% (alcohol level of water)
C = 300 hectoliters (total mixture)
Final Property:
c = ? (final alcohol percentage)
Calculation Steps:
Plug in known quantities:
200 × 7.15 + 0 × 100 = 300c
Isolate c:
1430 = 300c
c = 4.77% alcohol by volume
Note: Use de-aerated water in industrial processes to avoid increasing dissolved oxygen.
Example 2: Caustic Makeup Problem
Problem: How much 30% caustic must be added to raise causticity?
Given:
A = 200 hectoliters (initial solution)
a = 3%
B = ? (unknown volume of 30% caustic)
b = 30% (caustic concentration to add)
C = 4% (desired final concentration)
Calculation Steps:
Plug in values:
200 × 3 + 30B = (200 + B) × 4
Solve for B:
600 + 30B = 800 + 4B
26B = 200
B = 200/26 = 7.69 hectoliters
Knowledge Check
True or False: The mixing formula is a complex equation.
Answer: False
Reason: It is a simple algebraic formula.
Additional Problem: Mixing Nitric Acid
Problem: Volumes needed for 25% nitric acid solution.
Given percentage and strength, calculate needed volumes.
Answer: 52.6 liters is required.
Example 3: Peroxyacetic Acid Makeup Problem
Problem: How to make a 100 part per million solution from a 5.5% solution.
Given:
Total volume = 1 liter (1,000 milliliters)
a = 0 ppm (concentration in water)
B = ? (volume of peroxyacetic acid needed)
b = 55,000 ppm (concentration of the 5.5% solution)
C = 100 ppm (desired concentration)
Calculation Steps:
Convert units and set up equations:
0 × (1000 - B) = 0 (zero concentration)
55,000B = 100,000
Solve for B:
B = 1.8 milliliters of peroxyacetic acid.
Calculate the amount of water to add:
998.2 milliliters (1000 - 1.8)